#### To determine

Whether the series is convergent or divergent and obtain the sum if the series is convergent.

#### Answer

The series is convergent to the sum 2+2≈3.414.

#### Explanation

**Given:**

The series is ∑k=0∞(2)−k.

**Result used:**

The geometric series ∑n=1∞arn−1 (or) a+ar+ar2+⋯ is convergent if |r|<1 and its sum is a1−r, where *a* is the first term and *r* is the common ratio of the series.

**Calculation:**

The given series can be written as follows,

∑k=0∞(2)−k=∑k=0∞1(2)k=∑k=0∞(12)k=(12)0+(12)1+(12)2+(12)3+⋯=1+(12)+(12)2+(12)3+⋯

Here, the first term of the series is *a* = 1 and the common ratio of the series is,

r=121=12

The absolute value of *r* is,

|r|=|12|=12=0.707<1

Since |r|<1 and by using Result stated above, the series ∑k=0∞(2)−k is convergent.

Obtain the sum of the series.

Since a=1 and r=12.

∑k=0∞(2)−k=11−12=12−12=22−1

Multiply by the conjugate 2+12+1 and simplify the terms,

∑k=0∞(2)−k=22−1×2+12+1=2+22−1=2+2≈3.414

Thus, the sum of the series is 2+2≈3.414.

Therefore, the series is convergent to the sum 2+2≈3.414.